3.75X10^5 That seems to be the answer. You could type that in your calculator to see what you get.
Answer:
x = ± 
Step-by-step explanation:
To find the zeros equate the polynomial to zero, that is
x² - 7 = 0 ( add 7 to both sides )
x² = 7 ( take the square root of both sides )
x = ±
Thus the exact solutions are
x = -
, x = 
Answer:
(5,-2) or y=3(x-5)^2 -2
Step-by-step explanation:
The base case of
is trivially true, since

but I think the case of
may be a bit more convincing in this role. We have by the inclusion/exclusion principle

with equality if
.
Now assume the case of
is true, that

We want to use this to prove the claim for
, that

The I/EP tells us

and by the same argument as in the
case, this leads to

By the induction hypothesis, we have an upper bound for the probability of the union of the
through
. The result follows.
